Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 639: 67

$\dfrac{\pi^2}{8}$

$\Sigma_{k=1}^\infty \dfrac{1}{k^2}=\Sigma_{k=1}^\infty \dfrac{1}{2k^2}+\Sigma_{k=1}^\infty \dfrac{1}{(2k-1)^2}$ or, $\Sigma_{k=1}^\infty \dfrac{1}{k^2}=\dfrac{1}{4} \Sigma_{k=1}^\infty \dfrac{1}{k^2}+\Sigma_{k=1}^\infty \dfrac{1}{(2k-1)^2}$ or, $\dfrac{\pi^2}{6}=\dfrac{1}{4} \dfrac{\pi^2}{6}+\Sigma_{k=1}^{\infty} \dfrac{1}{(2k-1)^2}$ or, $\Sigma_{k=1}^{\infty} \dfrac{1}{(2k-1)^2}=\dfrac{\pi^2}{8}$